Dimensionality theorem irreps finite dimensional algebra For representations of finite groups it is known that the regular representation of a finite group is the direct sum of irreps, each with dimension equal to the multiplicity with which they appear. I assume the same does not hold for the algebra case since we can't really talk about conjugacy classes and the left action of an algebra on itself does not necessarily permute the basis elements of the algebra.
I do wonder whether there exists some (probably weaker) statement about the decomposition of the regular representation of irreps of a finite dimensional complex algebra?
 A: If $A$ is any ring (and in particular any finite-dimensional complex algebra) then the regular representation of $A$ on itself by left multiplication has submodules consisting of exactly the left ideals, more or less by definition. It is completely reducible iff $A$ is semisimple, again more or less by definition, and for $A$ finite-dimensional over a field (or more generally for $A$ Artinian) this occurs iff $A$ has vanishing Jacobson radical.
If $A$ is Artinian but does not have vanishing Jacobson radical then $A/J(A)$ is semisimple and is the maximal quotient with this property, and by the Artin-Wedderburn theorem it is a finite direct product of matrix rings $M_n(D)$ over division rings. In particular if $A$ is finite-dimensional over a field $k$ then $A/J(A)$ is a finite direct product of matrix rings over division $k$-algebras, so if $k$ is algebraically closed $A/J(A)$ is a finite direct product of matrix rings $M_n(k)$ over $k$.
In the case of semisimple algebras over an algebraically closed field $k$, each $d$-dimensional simple module $S$ corresponds to a matrix ring factor $M_d(k) = \text{End}_k(S)$ which, in terms of the left action of $A$ on itself consists of $d$ copies of $S$. A conceptual way to see this is that when $A$ is a semisimple module then the multiplicity of $S$ in $A$ is $\dim_k \text{Hom}_A(A, S)$, and we have $\text{Hom}_A(A, S) \cong S$ since $A$ is the free $A$-module. More generally, if $A$ is a semisimple ring and $S$ a simple module with endomorphism ring $\text{End}_A(S) \cong D$, corresponding to a matrix ring factor $M_d(D)$ in $A$, then the multiplicity of $S$ in $A$ is
$$\dim_D \text{Hom}_A(A, S) = \dim_D S = d.$$
So $S$ still occurs with multiplicity equal to its dimension but its dimension needs to be understood as a module over its endomorphism ring $D$ (which is a division ring by Schur's lemma).
